# The differentiability of convolutions

Yes, again, this type of question. Similar ones this and this.

I come with another variant. Let $f\in\mathcal{S}$, i.e. Schwartz function, and $g\in L^{p}(\mathbb{R}^d),p\in[1,\infty]$. The following should still hold $$\partial^\alpha(f*g) = (\partial^\alpha f)*g.$$

Basically we only need to prove the following case and the rest is simply by induction.

Here is my proof for $p\in[1,\infty)$, where are use the proposition $$D_i(f*g) = (D_if)*g,$$ provided $f,g\in\mathcal{S}$ and $D_i:=\frac {\partial}{\partial x_i}$.

Taking arbitrarily fixed $x$, \begin{align*} |D_i(f*g)(x)| &= \left|\lim_{h\rightarrow 0}\frac{f*g(x+he_i)-f*g(x)}{h}\right|\\ &\leq \lim_{h\rightarrow 0}\int_{\mathbb{R}^d}\left|\frac{f(x+he_i-y)-f(x-y)}{h}\right| |g(y)|\,\mathrm{d}y\\ &= \lim_{h\rightarrow 0}\int_{\mathbb{R}^d}|f_h(x-y)| |g(y)|\,\mathrm{d}y\\ &\leq \lim_{h\rightarrow 0}\||f_h(x-\cdot)|\|_{q}\|g\|_p\\ &= \|D_i f\|_{q}\|g\|_p. \end{align*} By taking supremum, we have shown that $D_i(f*g)$ is bounded. Hence $D_i(f*g)(x)$ is well-defined for all $x$.

Then, take a sequence $g_n\in\mathcal{S}$ such that $g_n\rightarrow g$ in $L_p$. Since \begin{align*} |D_i(f*g)-D_i(f*g_n)| &= \lim_{h\rightarrow 0}\left| \int_{\mathbb{R}^d}|f_h(x-y)| |g(y)-g_n(y)|\,\mathrm{d}y \right|\\ &\leq \lim_{h\rightarrow 0}\||f_h(x-\cdot)|\|_{q}\|g-g_n\|_p\\ &= \|D_i f\|_{q}\|g-g_n\|_p \rightarrow \quad \text{as}\quad n\rightarrow 0. \end{align*}

Besides, we know $$D_i (f*g_n) = (D_i f)*g_n,\quad \forall n\in\mathbb{N}.$$ Hence, \begin{align*} |D_i(f*g)-(D_i f)*g| &\leq |D_i(f*g)-(D_i f)*g - D_i (f*g_n) + (D_i f)*g_n|\\ &\leq |D_i(f*g)- D_i (f*g_n)| + |(D_i f)*g_n-(D_i f)*g|\\ &\leq |D_i(f*g)- D_i (f*g_n)| + \|(D_i f)\|_{q}\|g_n-g\|_{p}\rightarrow 0 \quad\text{as}\quad n\rightarrow \infty, \end{align*} where we applied Young's inequality in the last step.

Is it correct? And how should I work on the case $p=\infty$.

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The expression within the $||.||_q$ converges pointwise to $f'(x-y)$. And as you have an integrable majorant ($f$ and $f'$ decay faster than any polynomial), you can apply the dominated convergence theorem. – Vobo Jun 4 '13 at 18:51
@Vobo Thanks for the comment. But we only know $g\in L^p$, how can I handle this? – newbie Jun 4 '13 at 23:42
I am not completely sure about the well-definedness. For instance, $\lim_{n \to \infty} |(-1)^n| \le 1$, hence the sequence is bounded, but the limit does not exist. – Cloudscape Jul 22 '15 at 22:01

## 1 Answer

Your answer is correct. Let me show you the case $p=\infty$. As above, it is well-defined. Let $p(x)=(1+||x||^2)^n$. Note that $1/p$ is integrable and $pf$ and $p(D^if)$ are (uniformly) bounded. Then $$\Big| D^i (f\ast g) (x) - (D^i f\ast g)(x)\Big| \leq ||g||_\infty \lim_{h\to 0} \int |f_h(x-y) - (D^i f)(x-y)| dy.$$ As $f_h\to D^i f$ pointwise, you need to justify the dominated convergence theorem. This can be done by adding the factor $\frac{p(x-y)}{p(x-y)}$ to the integral: You still have $$\lim_{h\to 0} |p(x-y) f_h(x-y) - p(x-y)(D^i f)(x-y)| = 0$$ pointwise and this term is uniformly bounded. This bound times $1/p$ is an integrable majorant, so the use of Lebesgue is justified.

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Thanks. I also figured out that we can use the property of Schwartz function to argue it. – newbie Jun 5 '13 at 21:14